Problem: Simplify; express your answer in exponential form. Assume $r\neq 0, a\neq 0$. $\dfrac{{(r^{5})^{-1}}}{{(r^{-5}a^{-1})^{-5}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{5}}$ to the exponent ${-1}$ . Now ${5 \times -1 = -5}$ , so ${(r^{5})^{-1} = r^{-5}}$ In the denominator, we can use the distributive property of exponents. ${(r^{-5}a^{-1})^{-5} = (r^{-5})^{-5}(a^{-1})^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{5})^{-1}}}{{(r^{-5}a^{-1})^{-5}}} = \dfrac{{r^{-5}}}{{r^{25}a^{5}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-5}}}{{r^{25}a^{5}}} = \dfrac{{r^{-5}}}{{r^{25}}} \cdot \dfrac{{1}}{{a^{5}}} = r^{{-5} - {25}} \cdot a^{- {5}} = r^{-30}a^{-5}$.